2.3 Limits and continuity

Try a repeating pattern.

Many answers are possible; the tangent function behaves like this.

Many answers are possible; the tangent function behaves like this.

Compare what is given to you to the definition of continuity.

True.

True. Since $f(t)$ is continuous at $t=5$, that means $\ds\lim_{t \to 5} f(t)=f(5)$. For that to be true, $f(5)$ must exist — that is, 5 is in the domain of $f(x)$.

Compare what is given to you to the definition of continuity.

True.

True. Using the definition of continuity, $\ds\lim_{t \to 5} f(t)=f(5)=17$.

What if the function is discontinuous?

In general, false.

In general, false. If $f(t)$ is continuous at $t=5$, then $f(5)=17$; if $f(t)$ is discontinuous at $t=5$, then $f(5)$ either does not exist, or is a number other than 17.

An example of a function with $\ds\lim_{t \to 5}f(t)=17 \neq f(5)$ is $f(t)=\left{ \begin{array}{lcr} 17&,&t \neq 5\ 0&,&t=5 \end{array} \right.$, shown below.

What is $h(0)$?

$\ds\lim_{x \to 0^+} h(x)=0$

Since $f(x)$ and $g(x)$ are continuous at zero, and since $g^2(x)+1$ must be nonzero, then $h(x)$ is continuous at 0 as well. According to the definition of continuity, then $\ds\lim_{x \to 0}h(x)$ exists and is equal to $h(0)=\frac{0f(0)}{g^2(0)+1}=0$.

Since the limit $\ds\lim_{x \to 0}h(x)$ exists and is equal to zero, also the one-sided limit $\ds\lim_{x \to 0^+}h(x)$ exists and is equal to zero.

Use the definition of continuity.

$k=0$

Using the definition of continuity, we need $k=\displaystyle\lim_{x \rightarrow 0} f(x)$. Since the limit is blind to what actually happens to $f(x)$ at $x=0$, this is equivalent to $k=\displaystyle\lim_{x \rightarrow 0} x\sin\left(\frac{1}{x}\right)$. So if we find the limit, we solve the problem.

For any nonzero value of $x$, $-1 \le\sin\left(\frac{1}{x}\right)\le 1$. So if we multiply $x$ by $\sin\left(\frac{1}{x}\right)$, the magnitude (absolute value) of $x$ either stays the same or gets closer to 0. Since $x$ is already approaching 0, $\displaystyle\lim_{x \rightarrow 0} x\sin\left(\frac{1}{x}\right)=0$.

So, when $k=0$, the function is continuous at $x=0$.

Find the domain: when is the denominator zero?

$(-\infty, -1)\cup (-1,1) \cup (1,+\infty)$

$f(x)$ is a rational function and so is continuous except when its denominator is zero. That is, except when $x=1$ and $x=-1$.

When is the denominator zero? When is the argument of the square root negative?

$(-\infty, -1)\cup (1,+\infty)$

The function is continuous when $x^2-1> 0$, i.e. $(x-1)(x+1)> 0$, which yields the intervals $(-\infty, -1)\cup (1,+\infty)$.

When is the denominator zero? When is the argument of the square root negative?

The function is continuous except at $x=\pm \pi, \pm 3\pi, \pm 5\pi, \dots$.

The function $1/\sqrt{x}$ is continuous on $(0,+\infty)$ and the function $\cos(x) + 1$ is continuous everywhere.

So $1/\sqrt{\cos(x) + 1 }$ is continuous except when $\cos x=-1$. This happens when $x$ is an odd multiple of $\pi$. Hence the function is continuous except at $x=\pm \pi, \pm 3\pi, \pm 5\pi, \dots$.

There are infinitely many points where it is not continuous.

$x \neq n\pi,$ where $n$ is any integer

The function is continuous when $\sin(x)\neq 0$. That is, when $x$ is not an integer multiple of $\pi$.

$x=c$ is the important point.

$\pm 2$

The function is continuous for $x\ne c$ since each of those two branches are polynomials. So, the only question is whether the function is continuous at $x=c$; for this we need

We compute

So, we need $8-c^2=c^2$, which yields $c^2=4$, i.e. $c=-2$ or $c=2$.

The important place is $x=0$.

$c=1$

The function is continuous for $x \ne 0$ since $x^2+c$ and $\cos cx$ are continuous everywhere. It remains to check continuity at $x=0$. To do this we must check that the following three are equal.

Hence when $c=1$ we have the limits agree.

The important point is $x=c$.

$-1$, $4$

The function is continuous for $x\ne c$ since each of those two branches are defined by polynomials. Thus, the only question is whether the function is continuous at $x=c$. Furthermore,

and

For continunity we need both limits and the value to agree, so $f$ is continuous if and only if $c^2-4 = 3c$, that is if and only if

Factoring this as $(c-4)(c+1) = 0$ yields $c=-1$ or $c=+4$.

The important point is $x=2c$.

$c=1$, $c=-1$

The function is continuous for $x\ne c$ since each of those two branches are polynomials. So, the only question is whether the function is continuous at $x=c$; for this we need

We compute

So, we need $6-2c^2=4c^2$, which yields $c^2=1$, i.e. $c=-1$ or $c=1$.